However one needs to consider also unbounded operators. Such operators not only fail (\ref{eq-13.4.1}) but they are not defined everywhere.

Definition 3.
Consider a linear operator $L:D(L)\to \mathsf{H}$ where $D(L)$ is a linear subset in $\mathsf{H}$ (i.e. it is a linear subspace but we do not call it this way because it is not closed) which is dense in $\mathsf{H}$ (i.e. for each $u\in \mathsf{H}$ there exists a sequence $u_n \in D(L)$ converging to $u$ in $\mathsf{H}$). Then

$z\in \mathbb{C}$ belongs to the resolvent set of $L$ if $(L-z)^{-1}: \mathsf{H}\to D(L)$ exists and is a bounded operator: $(L-z)^{-1}(L-z)=I$, $(L-z)(L-z)^{-1}=I$.

$z\in \mathbb{C}$ belongs to the spectrum of $L$ if it does not belong to its resolvent set. We denote spectrum of $L$ as $\sigma (L)$.

Remark 2.

Resolvent set is always open (and spectrum is always closed) subset of $\mathbb{C}$;

If $L$ is self-adjoint then $\sigma (L)\subset \mathbb{R}$;

If $L$ is bounded then $\sigma (L)$ is a bounded set.

Classification

Not all points of the spectrum are born equal! From now on we consider only self-adjoint operators.

Definition 5.

$z$ is an eigenvalue if there exists $u\ne 0$ s.t. $(A-z)u=0$. Then $u$ is called eigenvector (or eigenfunction depending on context) and $\{u:\, (A-z)u=0\}$ is an eigenspace (corresponding to eigenvalue $z$). The dimension of the eigenspace is called a multiplicity of $z$. Eigenvalues of multiplicity $1$ are simple, eigenvalues of multiplicity $2$ are double, ... but there could be eignvalues of infinite multiplicity!

The set of all eigenvalues is called pure point spectrum;

Eigenvalues of the finite multiplicity which are isolated from the rest of the spectrum form a discrete spectrum; the rest of the spectrum is called essential spectrum.

Definition 6.
$z\in \mathbb{C}$ belongs to continuous spectrum if $z$ is not an eigenvalue and inverse operator $(L-z)^{-1}$ exists but is not a bounded operator (so its domain $D((L-z)^{-1}$ is dense).

Remark 3.
Continuous spectrum could be classified as well. The difference between absolutely continuous and singular continuous spectra will be illustrated but one can define also multiplicity of continuous spectrum as well. However one needs a Spectral Theorem to deal with these issues properly.

Spectrum: examples

Example 1.
Schrödinger operator
\begin{equation}
L=-\frac{1}{2}\Delta + V(x)
\label{eq-13.4.5}
\end{equation}
with potential $V(x)\to +\infty$ as $|x|\to \infty$ has a discrete spectrum: its eignevalues $E_n\to +\infty$ have finite multiplicities. In dimension $d=1$ all these eigenvalues are simple, not necessarily so as $d\ge 2$.

Example 2.
Consider Laplacian on 2-dimensional sphere which appears after separation of variables for Laplacian in $\mathbb{R}^3$ in spherical coordinates in Subsection 6.3.2.
Then $-\Delta$ has a spectrum $\{E_n= n(n+1): n=0,1,\ldots\}$; $E_n$ is an eigenvalue of multiplicity $(2n+1)$. Corresponding eigenfunctions are spherical harmonics. See Definition 8.1.1.

Example 5.
Schrödinger operator (\ref{eq-13.4.5}) with potential $V(x)\to 0$ as $|x|\to \infty$ has a continuous spectrum $[0,+\infty)$ but it can have a finite or infinite number of negative eignvalues $E_n<0$.

If $|V(x)|\le M(|x|+1)^{-m}$, $m>2$ the number of eigenvalues is finite.

Perturbing it by a potential $V(x)$, $V(x)\to 0$ as $|x|\to \infty$
\begin{equation}
L=\sum _{j=1}^3 \gamma^j (-i\partial_{x_j}) + m\gamma^0 +V(x) I, \qquad m>0
\label{eq-13.4.8}
\end{equation}
can add a finite or infinite number of eigenvalues in spectral gap $(-m,m)$. They can accumulate only to the borders of the spectral gap.

Example 8.
Consider Schrödinger operator (\ref{eq-13.4.5}) with periodic potential in $\mathbb{R}^d$: $V(x+a)=V(x)$ for all $a\in \Gamma$ where $\Gamma$ is a lattice of periods, see Definition 4.B.1. Then $L$ has a band spectrum.

Namely on the elementary cellDefinition 4.B.3 $\Omega$ consider operator $L(k)$ where $k\in \Omega^*$ is a quasimomentum; $L(k)$ is given by the same formula as $L$ but s defined on functions which are quasiperiodic with quasimomentum $k$. Its spectrum is discrete:
$\sigma (L(k))=\{E_n (k): n=1,2,\ldots\}$.

As dimension $d=1$ we can do better than this: $E_n(k)$ are increasing (decreasing) functions of $k$ on $(0,\pi/a)$ (where $a$ is the period) as $n$ is odd (respectively even) and \begin{equation}E_n^*:=\max _{k\in [0,\pi/a]} E_n(k)\le E_{(n+1)*}:=\min _{k\in [0,\pi/a]} E_{n+1}(k)\label{eq-13.4.12}\end{equation} and for generic potential $V(x)$ all inequalities are strict and all all spectral gaps $(E_n^*,E_{(n+1)*})$ are open.

As dimension $d\ge 2$ only finite number of spectral gaps could be open.

Perturbation of such operator $L$ by another potential $W(x)$, $W(x)\to 0$ as $|x|\to \infty$ could can add a finite or infinite number of eigenvalues in spectral gaps. They can accumulate only to the borders of the spectral gaps.

Example 9.
In the space $\ell^2(\mathbb{Z})$ (which is the space of sequences $u_n$, $n=\ldots, -2,-1,0, 1,2,\ldots$ such that $\|u\|^2:=\sum_{n=-\infty} ^{\infty}|u_n|^2<\infty$) consider almost Mathieu operator (which appears in the study of quantum Hall effect).
\begin{equation}
(Lu)_n =u_{n+1}+u_{n-1}+2\lambda \cos (2\pi (\theta +n\alpha))
\label{eq-13.4.13}
\end{equation}
with $|\lambda|= 1$. Assume that $\alpha$ is a Diophantine number (which means it is an irrational number which cannot be approximated well by rational numbers; almost all irrational numbers (including all algebraic like $\sqrt{2}$) are Diophantine).

Then the spectrum $\sigma(L)$ is continuous (no eigenvalues!) but it is singular continuous: for any $\varepsilon>0$ it can be covered by the infinite sequence of segments of the total length $<\varepsilon$. As an example of such set see Cantor set.

Remark 4.

Example 8 was completely investigated only in the end of the 20-th century.

Spectrum: explanations

Landau levels

Consider Example 3:
Schrödinger operator in 2D with a constant magnetic and no electric field
\begin{equation}
L=\frac{1}{2} (-i\partial_x -\frac{1}{2}B y)^2 +
\frac{1}{2} (-i\partial_y +\frac{1}{2}B y)^2
\label{eq-13.4.14}
\end{equation}
with $B>0$ (or $B<0$) has a pure point spectrum. For simplicity assume that $B>0$. We apply a gauge transformation which for Schrödinger operator means multiplying it from the left and right by $e^{i\hbar^{-1} \phi (x)}$ and $e^{-i\hbar^{-1} \phi (x)}$ respectively with a real-valued $\phi$ (which is an unitary transformation) and replaces $-i\hbar\nabla$ by $-i\hbar\nabla - (\nabla\phi)$ (which is equivalent to changing vector potential $\mathbf{A}(x)$ by $\nabla \phi$ which in turn does not change $\nabla \times \mathbf{A}$. Taking $\hbar=1$ and $\phi= \frac{1}{2}B xy$ we arrive to
\begin{equation*}
L'=\frac{1}{2} (-i\partial_x -B y)^2 +
\frac{1}{2} (-i\partial_y )^2;
\end{equation*}
then making Fourier transform by $x\mapsto \xi$ we get
\begin{equation*}
L''=\frac{1}{2} (-\xi -B y)^2 +
\frac{1}{2} (-i\partial_y )^2;
\end{equation*}
and plugging $y=B^{-\frac{1}{2}} (y_{\textsf{new}} -B^{-1}\xi)$ we get
\begin{equation*}
\frac{1}{2} B (-\partial_y^2 +y^2)
\end{equation*}
which is a harmonic oscillator multiplied by $B$ and in virtue of Section 4.C its spectrum consists of eigenvalues
$E_n = |B|(n+\frac{1}{2})$, $n=0,1,2,\ldots$ which are called Landau levels.

However there is a "hidden variable" $\xi$, so eigenfunctions Hermite functions of $y$ but multiplied by arbitrary functions $C(\xi)$ rather than by constants which implies that these eigenvalues have constant multiplicities.

Band spectrum

Consider Example 8: Schrödinger operator with periodic potential in $\mathbb{R}^d$: $V(x+a)=V(x)$ for all $a\in \Gamma$ where $\Gamma$ is a lattice of periods.

In advanced Real Analysis it would be a decomposition of our Hilbert space $\mathsf{H}=L^2(\mathbb{R}^n)$ into direct integral of Hilbert spaces $\mathsf{H}(\mathbf{k})$ of such functions, and our operator is acting in each of those spaces separately, with a spectrum $\sigma (L(\mathbf{k}))=\{E_n (\mathbf{k}): n=1,2,\ldots\}$. This implies that $L$ has a band spectrum: it consists of spectral bands $\sigma_n:=[\min _{k\in \Omega^*} E_n(k) ,\max _{k\in \Omega^*} E_n(k)]$:
\begin{equation*}
\sigma(L) =\bigcup_{n=1}^\infty \sigma_n;
\end{equation*}
these spectral bands can overlap. On can prove that $E_n (\mathbf{k})$ really depend on $\mathbf{k}$ and are not taking the same value on some set of non–zero measure (another notion from Real Analysis) which implies that the spectrum $\sigma(L)$ is continuos.